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Byju's Answer

Standard XII

Physics

Charging of Capacitors

A parallel pl...

Question

A parallel plate capacitor has plates with area $A$ & separation $d$ . A battery charges the plates to a potential difference of $V_{0}$ . The battery is then disconnected & a di-electric slab of constant $K$ & thickness $d$ is introduced. Calculate the positive work done by the system (capacitor + slab) on the man who introduces the slab.

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Solution

charges $Q = C V = \frac{ε_{o} A}{d} V_{o}$
work sone= Initial energy-Final energy
$W = \frac{1}{2} C_{o l d} V^{2} - \frac{Q^{2}}{2 C_{n e w}}$
$= \frac{1}{2} (\frac{ε_{o} A}{d}) V_{o}^{2} - \frac{A ε_{o} V_{o}^{2}}{2 K d}$
$= \frac{ε_{o} A V_{o}^{2}}{2 d} [1 - \frac{1}{K}]$

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charges Q=CV=εoAdVowork sone= Initial energy-Final energyW=12ColdV2−Q22Cnew=12(εoAd)V2o−AεoV2o2Kd=εoAV2o2d[1−1K]

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work sone= Initial energy-Final energy
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$= \frac{ε_{o} A V_{o}^{2}}{2 d} [1 - \frac{1}{K}]$